Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef complex<double> point;
#define xx real()
#define yy imag()

#define REP(i, a, b) for(int i = (a); i < (int)(b); i++)
#define REPN(i, a, b) for(int i = (a); i <= (int)(b); i++)
#define FA(it, x) for(__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)
#define SZ(x) (int)(x).size()
#define BE(x) (x).begin(), (x).end()
#define SORT(x) sort(BE(x))
#define _1 first
#define _2 second

#define x1 gray_cat_x1
#define y1 gray_cat_y1

template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

#define file "I1"

const double EPS = 1e-9;
const double PI = acos(-1.);
const int INF = 1e9;
const ll MOD = 1e9 + 7;

const int MAXN = 2e5 + 5;

int p_inv[MAXN];

void solve(){
	int n, t, f, a;
	scanf("%d", &n);
	REPN(i, 1, n) {
		scanf("%d", &a);
		p_inv[a] = i;
	}
	t = f = 0;
	for(int i = n - 1; i >= 1; i--) {
		if (t == 0) {
			if (p_inv[i] > p_inv[i + 1]) {
				t = 1;
				f = i + 1;
			}
		} else {
			if (!(p_inv[f] < p_inv[i] && p_inv[i] < p_inv[i + 1]
				|| p_inv[i] < p_inv[i + 1] && p_inv[i + 1] < p_inv[f]
				|| p_inv[i + 1] < p_inv[f] && p_inv[f] < p_inv[i])) {
					t++;
					f = i + 1;
				}
		}
	}
	printf("%d\n", t);
}   

int main(){

    //freopen(file".in", "r", stdin); freopen(file".out", "w", stdout);
    int t = 1;
    //cin >> t;
    while(t--){
        solve();    
    }
}